1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another vector problem -- A ball suspended from 3 wires

  1. Nov 17, 2014 #1
    Question

    A ball with a mass 4 kg is suspended from 3 light in-extensible wires (diagram given)
    Given that the ball is in equilibrium, calculate the tension in each of the wires.

    My attempt

    T1 is equal to the weight of the ball which is 4g
    so T1 = 4g

    From there I dont know what to do but for T3 is the vertical component equal to 4g and to find T3 you can do 4gcos(70)= T3?

    is the above for T3 right? If not how do I do it?

    Not sure how to do T2 could someone please explain?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 17, 2014 #2
    Work with the horizontal components first. We know the horizontal components of T2 and T3 must be equal in magnitude. You can then relate T2 and T3.
     
  4. Nov 17, 2014 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You know the angles the tension forces make to each other, and you know one magnitude.
    You know how to add vectors head-to-tail ... and you know what shape these vectors have to make when you do that.
    You know some rules for finding lengths of sides when you know the angles.
    (Or split everything into horizontal and vertical components...)
     
  5. Nov 17, 2014 #4
    Thank you for the help :)

    so horizontal of T2=T2cos60
    horizontal of T3= T3sin70
    so T2cos60=T3sin70 (is this right?)
     
  6. Nov 17, 2014 #5

    jtbell

    User Avatar

    Staff: Mentor

    OK so far. Now, can you come up with a similar equation using the vertical components of T1, T2 and T3?
     
  7. Nov 17, 2014 #6
    T3cos(70)= vertical of T3
    T2sin(60)= vertical of T2
    T1=4g

    Does this mean that for the verticals of the tensions 4g=T3cos(70) + T2sin(60)?
     
  8. Nov 17, 2014 #7

    jtbell

    User Avatar

    Staff: Mentor

    Not quite. T2 is upwards. T1 and T3 are downwards.
     
  9. Nov 17, 2014 #8
    Okay so the verticals of T1 and T3 must equal the vertical of T2?
     
  10. Nov 17, 2014 #9

    jtbell

    User Avatar

    Staff: Mentor

    Correct.

    With horizontal components, the total to the left equals the total to the right.
    With vertical components, the total upwards equals the total downwards.

    (for a system in equlibrium, like this one.)
     
  11. Nov 17, 2014 #10
    I dont understand what to do after that because substituting in the equations means the tensions cancel out of the equation?
     
  12. Nov 17, 2014 #11

    jtbell

    User Avatar

    Staff: Mentor

    You have two equations in two unknowns, T2 and T3. They should not cancel out when you solve the equations together. If they do, you've made a mistake in your algebra. Can you show your work?
     
  13. Nov 17, 2014 #12
    Starting equations:
    T2cos(60)=T3sin(70)
    T2(sin60) = T3cos(70) + 4g

    Then do I rearrange one of them to make one of the T's the subject and sub it into the other equation?


    Thank you for all the help btw :)
     
  14. Nov 17, 2014 #13

    jtbell

    User Avatar

    Staff: Mentor

    Exactly. :D
     
  15. Nov 17, 2014 #14
    Would you be able to give me an example of how to rearrange and substitute these equations to find one of the tensions? as im having trouble doing this
     
  16. Nov 17, 2014 #15

    jtbell

    User Avatar

    Staff: Mentor

    Show us what you've tried and where you get "stuck" and I or someone else can probably help you get "unstuck."
     
  17. Nov 17, 2014 #16
    Starting equations:
    T2cos(60)=T3sin(70) (Equation 1)
    T2(sin60) = T3cos(70) + 4g (Equation 2)

    Rearranging T2(sin60) = T3cos(70) + 4g to get T2 = (T3cos(70) + 4g)/sin(60) then substitue it into equation 1

    this gives (T3cos(70) + 4g)/sin(60)xcos(60) = T3sin(70)

    then divided both sides by cos(60) to give

    (T3cos(70) + 4g)/sin(60) = (T3sin(70))/cos60

    Times both sides by (sin60)

    (T3cos(70) + 4g = [(T3sin(70))/cos60]x sin60

    this gives in numerical form

    0.3420201433T3 + 39.24 = 1.627595363T3

    Then subtract 0.3420201433T3 from both sides

    39.24 = 1.28557522T3

    then divde both sides by 1.28557522

    T3 = 30.52N

    for T3 this is what I got, is it correct?
     
  18. Nov 17, 2014 #17

    jtbell

    User Avatar

    Staff: Mentor

    I'm starting to look over your algebra, but you can check the final answers yourself. Use your value of T3 to find T2, then substitute both into your starting equations (at the top of your post) and see if they work out.
     
  19. Nov 17, 2014 #18

    jtbell

    User Avatar

    Staff: Mentor

    It looks OK to me so far. I wasn't sure at first because in your first post I missed the mass of the ball being "4 kg" and assumed that by "4g" in your equations you meant 4 grams. :oops: So I got a different number for T3 at first. I fixed that and now we agree.
     
  20. Nov 17, 2014 #19
    yeah sorry i mean 4 times by gravity when saying 4g
    Okay thank you for all the help :D

    to find T2 is it alright just to sub in my value of T3 into one of my first equations to get its value?

    Also is there a rep system on here so I can rep you? (new to this forum)
     
  21. Nov 17, 2014 #20

    jtbell

    User Avatar

    Staff: Mentor

    Right. Either equation should work. If you try both of them and get the same value for T2, that will also serve as a check.

    Not the kind that I think you're thinking of. We have other ways of identifying the "good guys."
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Another vector problem -- A ball suspended from 3 wires
  1. Suspended ball (Replies: 2)

  2. Suspended Ball (Replies: 2)

Loading...